document.write( "Question 272272: I am having extreme difficulty with this section of my homework. I have to find a polynomial function of least degree having only real coefficients with zeros.
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document.write( "The book tells me that f(x)=x-(1+i) has 1+i as a zero, but the conjugate 1-i is not a zero. Why? If I use the synthetic division (as the book says) to figure out if f(k)=0, I get don't understand why 1-i is not a zero. All I keep getting is the same equation by using synthetic division.\r
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document.write( "The problem I am having to do it:\r
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document.write( "2+, 2-
, and 2+3i.\r
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document.write( "By not understanding why why 1-i is not a zero, I am confused as to what is a zero in the above equation.\r
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document.write( "I THINK that 2+ and 2+3i are going to be zeros for the problem, giving me an equation that looks something like:\r
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document.write( "f(x)= [x-(2+)][x-(2-
)][x-(2+3i)][x-(2-3i)]\r
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document.write( "If someone could please help me with this I would GREATLY appreciate it. \n" );
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Algebra.Com's Answer #199234 by edjones(8007)![]() ![]() You can put this solution on YOUR website! x-(1+i)=0 \n" ); document.write( "let 1+i=a so you will understand better \n" ); document.write( "x-a=0 \n" ); document.write( "x=a \n" ); document.write( "x=1+i piece of cake. you are over-thinking this one. \n" ); document.write( "A first degree equation has only one zero. \n" ); document.write( ". \n" ); document.write( "Ed \n" ); document.write( " |